Chapter 10: Problem 40
Find an equation for the set of points in an xy-plane that are equidistant from the point \(P\) and the line \(L\) $$P(5,-2) ; \quad k y=4$$
Short Answer
Expert verified
The equation is \( x^2 - 10x + 12y = -13 \).
Step by step solution
01
Understanding the Components
We need to find an equation for the set of points that are equidistant from the point \( P(5, -2) \) and the line \( L: y = 4 \). This implies that any point \( (x, y) \) on this set is the same distance from both \( P \) and \( L \).
02
Distance Formula from a Point
The distance between a point \( (x, y) \) and \( P(5, -2) \) is given by the formula \( d = \sqrt{(x-5)^2 + (y + 2)^2} \).
03
Distance Formula from a Line
The distance from a point \( (x, y) \) to the line \( y = 4 \) can be calculated using the formula \( d = |y - 4| \). This is the vertical distance between the point and the horizontal line.
04
Setting Up the Equation
Since points \( (x, y) \) should be equidistant from \( P \) and the line, set the distances equal: \[ \sqrt{(x-5)^2 + (y+2)^2} = |y - 4| \]
05
Eliminating the Square Root and Absolute
Square both sides to eliminate the square root: \[ (x-5)^2 + (y+2)^2 = (y-4)^2 \] Expand and simplify both sides of the equation.
06
Expand and Simplify
Expanding the equation we get: \[ (x-5)^2 + (y+2)^2 = y^2 - 8y + 16 \] Expanding further gives: \[ x^2 - 10x + 25 + y^2 + 4y + 4 = y^2 - 8y + 16 \] Cancel \( y^2 \) from both sides, and rearrange the terms: \[ x^2 - 10x + 29 + 12y = 16 \]
07
Solving for y
Finally, simplify the equation for convenience which results in: \[ x^2 - 10x + 12y = -13 \]
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Distance Formula
The distance formula is a crucial tool to calculate how far apart two points are in a coordinate plane, often the xy-plane. It is derived from the Pythagorean theorem and is used when the coordinates of two points are known. For two points, \(A(x_1, y_1)\) and \(B(x_2, y_2)\), the distance \(d\) between them is calculated as follows:
- \(d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)
- \(d = \sqrt{(x-5)^2 + (y+2)^2}\)
XY-Plane
The xy-plane is a two-dimensional space where each point is determined by a pair of numerical coordinates. These coordinates are often written as \( (x, y) \), with \( x \) representing the horizontal position, and \( y \) representing the vertical position.
- The horizontal axis is called the x-axis.
- The vertical axis is called the y-axis.
- The point where both axes meet is called the origin, given by \( (0,0) \).
Geometric Locus
In geometry, a geometric locus is a set of points that satisfy certain conditions or rules. These points could form lines, curves, or shapes based on the set criteria. Imagine a geometric locus like a path or a boundary where points have a specific relationship to other objects.
- In the given exercise, we want our geometric locus (call it line or curve) to have all points equidistant from a fixed point and a line.
- The established condition led us to equal the distance formulas, resulting in a curve that satisfies both conditions.
Horizontal Line
A horizontal line in the xy-plane is one that runs parallel to the x-axis and maintains a constant y-coordinate for all its points. In simpler terms, it means all points on that line have the same height above or below the x-axis.
- The equation for a horizontal line is \( y = c \), where \(c\) is a constant, in this case, \y = 4\.
- In calculating distances, the use of the absolute difference between the y-values gives the vertical distance from a point to the line.